Using the chern character, it can be shown that there is no complex structure on $S^n$ if $n > 6$: See May's book: if $S^{2n}$ has a complex structure, let $\tau$ be the tangent bundle. $c_n(\tau) = \chi(S^{2n}) = 2$ must be divisible by $(n-1)!$ by Husemöller, Fibre bundles, chapter 20, Theorem 9.8. So the only case left is $S^4$ and $S^6$.

1 Answer
1

It is known that $S^4$ doesn't even have an almost complex structure, and the case for $S^6$ is open. The lack of almost complex number can be proved a number of ways, one way is by showing that an almost complex, compact, four manifold with $\dim_{\mathbb{Q}}H^2(X,\mathbb{Q})=0$ has $\chi(X)=0$, but the four sphere doesn't. (It follows from the index theorem, here's a quick reference, first result.)

A clarifying comment: $S^6$ DEFINITELY has an almost complex structure, but it's not integrable. It is open if there is a complex structure, and for more details, see the other question that Reid linked to.
–
Charles SiegelJan 13 '10 at 17:11